Question

displacement x = tcube - 12t + 10, the acceleration of particle when velocity is zero is given by 1. 12units 2.2units 3. 10 units 4. 1 units​

Answer 1

Correct Question

Displacement of a particle is given as x(t) = t³ - 12t + 10,the acceleration of the particle when velocity is zero is :

(A) 12 m/s² ✓

(B) 2 m/s²

(C) 10 m/s²

(D) 1 m/s²

Solution

The correct option is (A) 12 m/s²

Given

The position of the particle is given as :

$\sf \: x(t) = {t}^{3} - 12t + 10$

To finD

Acceleration Of The Particle

$\rule{300}{1}$

Differentiating x w.r.t to t,we get velocity of the particle :

$\sf \: v(t) = \dfrac{dx}{dt} \\ \\ \longrightarrow \: \sf \: v(t) = \dfrac{d( {t}^{3} - 12t + 10)}{dt} \\ \\ \longrightarrow \: \underline{ \boxed{\sf \: v(t) =( 3 {t}^{2} - 12 )\: \: {ms}^{ - 1} }}$

The velocity of the particle is zero

$\colon \implies\sf \: 3 {t}^{2} - 12 = 0 \\ \\ \colon \: \implies \: \sf \: 3 {t}^{2} = 12 \\ \\ \colon \: \implies \: \sf {t}^{2} = 4 \\ \\ \colon \: \implies \: \sf \: t = \pm \sqrt{4}$

Time is always positive

$\large{ \colon \: \implies \: \boxed{ \boxed{\sf \: t = 2s}}}$

$\rule{300}{1}$

Now,

Differentiating v w.r.t t,we get acceleration :

$\sf \: a(t) = \dfrac{dv}{dt} \\ \\ \longmapsto \: \sf \: a(t) = \dfrac{d( {3t}^{2} - 12)}{dt} \\ \\ \longmapsto \: \underline{ \boxed{\sf \: a(t) = 6 {t}^{} \: \: {ms}^{ - 2} }}$

When t = 2s,

$\huge{ \longmapsto \: \boxed { \boxed{ \sf \: a(t) = 12 \: {ms}^{ - 2} }}}$

$\rule{300}{1}$

$\rule{300}{1}$

Answer 2

$\Large{\underline{\underline{\green{\mathfrak{Given :}}}}}$

• x(t) = t³ - 12t + 10

• Velocity (v) is 0

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$\Large{\underline{\underline{\blue{\mathfrak{To \: Find :}}}}}$

• Acceleration of Particle

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$\Large{\underline{\underline{\red{\mathfrak{Solution :}}}}}$

As we are given that,

$\large{\underline{\boxed{\sf{x(t) \: = \: t^3 \: - \: 12t \: + \: 10}}}} \\ \\ \small{\underline{\sf{\gray{\dag \: \: \: \: \: \: Differentiate \: x \: with \: respect \: to \: t \: \: \: \: \: \: \dag}}}} \\ \\ \implies {\sf{v(t) \: = \: \dfrac{d(t^3 \: - \: 12t \: + \: 10)}{dt}}} \\ \\ \implies {\sf{v(t) \: = \: 3t^2 \: - \: 12}} \\ \\ {\huge{\underline{\boxed{\sf{v(t) \: = \: 3t^2 \: - \: 12}}}}}$

___________________________

$\small{\underline{\gray{\sf{\dag \: \: \: \: \: \: As \: velocity \: of \: particle \: is \: zero \: \: \: \: \: \: \dag}}}} \\ \\ \implies {\sf{0 \: = \: 3t^2 \: - \: 12}} \\ \\ \implies {\sf{3t^2 \: = \: 12}} \\ \\ \implies {\sf{t^2 \: = \: \dfrac{12}{3}}} \\ \\ \implies {\sf{t^2 \: = \: 4}} \\ \\ \implies {\sf{t \: = \: \sqrt{4}}} \\ \\ \implies {\sf{t \: = \: 2s}} \\ \\ \huge{\underline{\boxed{\sf{t \: = \: 2s}}}}$

___________________________

$\small{\underline{\gray{\sf{\dag \: \: \: \: \: \: Differentiate \: v \: with \: respect \: to \: t \: \: \: \: \: \: \dag}}}} \\ \\ \implies {\sf{a(t) \: = \: \dfrac{d(3t^2 \: - \: 12)}{dt}}} \\ \\ \implies {\sf{a(t) \: = \: 6t}} \\ \\ \implies {\sf{a(t) \: = \: 6 \: \times \: 2}} \\ \\ \implies {\sf{a(t) \: = \: 12 \: ms^{-2}}} \\ \\ \huge{\underline{\boxed{\sf{a(t) \: = \: 12 \: ms^{-2}}}}}$